解:(1)由S1=a1=1,S2=1+a2,得3t(1+a2)﹣(2t+3)=3t,
![](http://img.shitiku.com.cn/uploads/allimg/20191014/20191014190917-33075.png) ![](http://img.shitiku.com.cn/uploads/allimg/20191014/20191014190917-50135.png) 又3tSn﹣(2t+3)Sn﹣1=3t, 3tSn﹣1﹣(2t+3)Sn﹣2=3t(n=3,4,) 两式相减,得:3tan﹣(2t+3)an﹣1=0,
![](http://img.shitiku.com.cn/uploads/allimg/20191014/20191014190917-97375.png) (n=3,4,) 综上,数列{an}为首项为1,公比为 的等比数列 (2)由 ,得 , 所以{bn}是首项为1,公差为 的等差数列,
b1b2﹣b2b3+b3b4﹣b4b5+…+b2n﹣1b2n﹣b2nb2n+1 =(b1﹣b3)b2+(b3﹣b5)b4+…+(b2n﹣1﹣b2n+1)b2n =![](http://img.shitiku.com.cn/uploads/allimg/20191014/20191014190920-52629.png) =![](http://img.shitiku.com.cn/uploads/allimg/20191014/20191014190920-34829.png) |